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Thread: Flow capacity of pipe

  1. #1

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    Flow capacity of pipe

    I have a scenario where there is a reservoir (750 BBL) with 4 feet of head discharging through a 10 inch clean PVC (100 foot long) with 5 elbows and a strainer. I have been trying to model the maximum flow rate through the pipe but its proving unsuccessful. The pipe is level from the base of the tank to the discharge point. Any ideas or suggestions?

  2. #2
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    David,

    Quick Questions

    1) How exactly are you modeling this and with what software package are you using?

    2) Is this pipe discharging to atmosphere?
    -> If yes, you would generally not use a demand to represent the discharge, but rather either an emitter or a "reservoir" set to match the HGL of the discharge. An emitter on a junction is the best way to model a free discharge. Using a very large emitter coef like 10K or 20K will show the maximum flow possible, but your actual discharge may be less. The InfoWater UDF User Guide can be found in this directory on your computer -> C:\Program Files (x86)\InfoWater\Help as the InfoWater UDF User Guide.pdf and has a discussion of how to calculate an emitter coefficient based on the nozzle size, although it is limited to US units.

    Here is an excerpt of this page as it may explain a few things that may help you better understand emitters.
    (click if need larger image)
    UDF User Guide P 22 -Emitters.jpg

    2) Is the pipe flowing full at the discharge point?
    -> InfoWater which uses EPANET assumes all calculations that the pipes are flowing full. If the pipe is known to run less than 100% full, you will need to use a gravity flow based program like InfoSewer, or InfoSWMM to calculate flow where the pipes are not completely full.

    Other thoughts:

    InfoWater pipe flows are based on 2 primary things. 1) What is the head available to drive the flow (Water head at known locations must be satisfied) and 2) All demands must be satisfied. This is why a pipe break or a free discharge is difficult to model as a demand and is best modeled as an emitter or even a reservoir set at the HGL of the discharge (which is actually what EPANET uses for an emitter, with the resistance of flow to the reservoir based on the emitter coefficient). So to model a free discharge correctly the heads must match what is occurring, and the flow requirements form demands or a free discharge must be modeled correctly for it to replicate what is occurring in the real system.

    Please reply if you have any additional questions or comments we can assist you with.

    Patrick Moore

  3. #3

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    The pipe is discharging into a pond below it.

  4. #4

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    Sorry I'm using the Infowater package

  5. #5
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    David, I would model the discharge as a "reservoir" or even a "tank" as the pond has a known volume and depth. I would suspect something is restricting the flow into the pond somehow or refilling the upper tank as the head difference of 4 ft is enough to drain the upper "tank" given enough time, but as long as you replicate the pipes and their headloss between and have the head difference adequately modeled this should replicate well.

    How are you modeling this and then "comparing" it to the real world?

    The headloss between the upper tank and the lower tank would be induced by what flow would be needed to account for the head differences between the tanks. Your strainer may be providing more headloss than you expect especially if it get's clogged at all which may also reduce the flow you actually "see" in the pipe compared to the model. If this is not accounted for in the model your model flows would likely exceed the real world flows. The model flow would be based on the flow required to make the HGL's between the tanks match by inducing flow that would cause exactly 4 ft of headloss between the tanks.

    Patrick Moore

  6. #6

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    What I did was set the 750 BBL tank as a "Tank" and set the outfall at 100 feet away. The 10" pipe is level throughout and at the base of the Tank to the outfall pond.

  7. #7
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    David, I contacted you via email so we can run a quick WebEx to resolve your remaining questions.

    Patrick Moore

  8. #8
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    Just of the benefit of the Forum, let me post the resolution we found after we had a call.

    1) Your model needed at least one junction in it for the model to solve
    2) The head of the starting tank will equal the tank elevation plus the initial level in your case this was 100 ft + 4 ft for a total head of 104 ft.
    3) The receiving head on the reservoir was adjusted form 0 ft to 100 ft so that there was only 4 ft of head difference between the tank and reservoir.
    4) Minor losses if desired can also be added to the pipe to induce additional headloss as well.
    5) Running the model calculated the flow properly once these missing factors were added.

    Patrick Moore

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